If a-0- then -a-a- -A- 1-2-4 a-1B- 1-2-1-4 a-1-C- 1-2-1-4 a-1-D- None of these

The correct option is B x = √(a + √(a + √(a + √(a + ......∞)))) ⇒ x = √((a+x)) ⇒ x^2 x a = 0 ⇒ x = 1/2 [ 1 ± √((1 + 4a))] ∴ x = 1/2 [ 1 ± √((1 + 4a) ...

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Standard XII

Mathematics

Quadratic Formula for Finding Roots

If a>0, then ...

Question

If a > 0, then $\sqrt{{a + \sqrt{(a + . . . .)}}}$ =

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None of these

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Solution

The correct option is B

x = $\sqrt{a + \sqrt{a + \sqrt{a + \sqrt{a + . . . . . . \infty}}}}$

$\Rightarrow$ x = $\sqrt{(a + x)}$ $\Rightarrow$ $x^{2} - x - a$ = 0

$\Rightarrow$ x = $\frac{1}{2} [1 \pm \sqrt{(1 + 4 a)}]$

$∴$ x = $\frac{1}{2} [1 \pm \sqrt{(1 + 4 a)}]$ .

Rejecting other value of x which is negative while x must be positive.

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If a > 0, then √{a+√(a+....)}=

The correct option is B x = √a+√a+√a+√a+......∞ ⇒ x = √(a+x) ⇒ x2−x−a = 0 ⇒ x = 12[1±√(1+4a)] ∴ x = 12[1±√(1+4a)] . Rejecting other value of x which is negative while x must be positive.

If a > 0, then $\sqrt{{a + \sqrt{(a + . . . .)}}}$ =